Compounds like CCL2F2 are known as chloroflourocarbons, or CFCs. These compounds were once widely used as refrigerants but are now being replaced by compounds that are believed to be less harmful to the environment. The heat of vaporization of CCl2F2 is 289J/g. What mass of this substance must evaporate in order to freeze 100g of water initially at 18 C. (The heat of fusion of water is 334J/g; the specific heat of water is 4.18J/g-K).
gah. help??
How much heat must be lost to move H2O from 18 to 0 degrees C?
mass x specific heat x delta T =
100g x 4.186 J/g*C x 18 = ??
How much heat must bae lost to freeze the water?
100g x 334 J/g = ??
Total heat that must be lost (pulled out of the water) is the sum of the two.
How much heat can the CCl2F2 remove?
mass x 289 J/g = ??
Can you take it from here?
Sure, I can help! Let's break it down step by step.
First, we need to calculate how much heat is required to cool the water from 18°C to 0°C using the formula:
heat = mass x specific heat x delta T
where mass = 100g, specific heat = 4.18 J/g-K, and delta T = 18°C - 0°C = 18°C:
heat = 100g x 4.18 J/g-K x 18°C = 7536 J
Next, we need to determine the amount of heat required to freeze the water. The heat of fusion of water is given as 334 J/g, so:
heat = 100g x 334 J/g = 33400 J
The total amount of heat that needs to be removed from the water is the sum of the two values:
total heat = 7536 J + 33400 J = 40936 J
Now, let's calculate how much heat the CCl2F2 can remove using its heat of vaporization, which is given as 289 J/g:
mass of CCl2F2 = total heat / heat of vaporization
mass of CCl2F2 = 40936 J / 289 J/g = 141.87 g
So, approximately 141.87 grams of CCl2F2 need to evaporate in order to freeze 100g of water initially at 18°C.
I hope that helps! Let me know if you have any more questions or need further assistance.
Yes, I can help you with the remaining steps.
To find the mass of CCl2F2 that must evaporate, we need to calculate the total amount of heat that needs to be removed from the water.
First, calculate the heat required to move from 18°C to 0°C:
mass x specific heat x delta T = 100g x 4.18 J/g°C x (18°C - 0°C) = 7524 J
Next, calculate the heat required to freeze the water:
100g x 334 J/g = 33400 J
Now, calculate the total heat that needs to be removed from the water:
7524 J + 33400 J = 40924 J
Since the heat of vaporization of CCl2F2 is 289 J/g, we can now determine the mass of CCl2F2 that must evaporate:
mass = total heat / heat of vaporization
mass = 40924 J / 289 J/g ≈ 141.7 g
Therefore, approximately 141.7 grams of CCl2F2 must evaporate to freeze 100 grams of water initially at 18°C.
Yes! I can help you continue from here.
To find out how much heat must be lost to move the water from 18 to 0 degrees Celsius, we can use the formula:
Heat = mass * specific heat * change in temperature
Plugging in the values, we get:
Heat = 100g * 4.18 J/g-K * (0 - 18)°C
= 100g * 4.18 J/g-K * (-18)°C
= -7536 J
The negative sign indicates heat loss.
Next, let's find out how much heat must be lost to freeze the water. We can use the formula:
Heat = mass * heat of fusion
Plugging in the values, we get:
Heat = 100g * 334 J/g
= 33400 J
Now, let's calculate the total heat that must be lost, which is the sum of the two values we just calculated:
Total Heat = -7536 J + 33400 J
= 25864 J
So, the total heat that must be lost to freeze 100g of water is 25864 J.
Now, we need to find out how much heat can be removed by the CCl2F2. We can use the formula:
Heat = mass * heat of vaporization
Let's plug in the value for the heat of vaporization of CCl2F2, which is 289 J/g:
Heat = mass * 289 J/g
Now, we need to find the mass of CCl2F2 that can remove the total heat of 25864 J. We can rearrange the formula to solve for mass:
mass = Heat / (heat of vaporization)
mass = 25864 J / 289 J/g
= 89.57 g
Therefore, the mass of CCl2F2 that must evaporate in order to freeze 100g of water initially at 18°C is approximately 89.57 grams.
I hope this helps! Let me know if you have any further questions.